1. **State the problem:** Water flows from a rectangular reservoir of length 10 m and width 6 m into a cylindrical tank of radius 3 m. The water depth in the reservoir decreases at a rate of 0.05 m/min. We need to find how fast the water level is rising in the cylindrical tank. 2. **Define variables:** Let $h_r$ be the water height in the rectangular reservoir (m), and $h_c$ be the water height in the cylindrical tank (m). 3. **Given data:** - Reservoir length $L=10$m - Reservoir width $W=6$m - Rate of change in reservoir height $\frac{dh_r}{dt} = -0.05$ m/min (negative because height is decreasing) - Radius of cylinder $r=3$m 4. **Volumes in reservoir and tank:** - Volume of water in reservoir $V_r = L \times W \times h_r = 60 h_r$ - Volume of water in cylindrical tank $V_c = \pi r^2 h_c = 9\pi h_c$ 5. **Since water flows from reservoir to tank, volume lost by reservoir equals volume gained by tank:** $$\frac{dV_r}{dt} = -\frac{dV_c}{dt}$$ 6. **Relate rates of change of volumes to heights:** $$\frac{dV_r}{dt} = 60 \frac{dh_r}{dt}$$ $$\frac{dV_c}{dt} = 9\pi \frac{dh_c}{dt}$$ 7. **Set up equation:** $$60 \frac{dh_r}{dt} = -9\pi \frac{dh_c}{dt}$$ 8. **Solve for $\frac{dh_c}{dt}$:** $$\frac{dh_c}{dt} = -\frac{60}{9\pi} \frac{dh_r}{dt}$$ Substitute $\frac{dh_r}{dt} = -0.05$: $$\frac{dh_c}{dt} = -\frac{60}{9\pi} (-0.05) = \frac{60 \times 0.05}{9\pi} = \frac{3}{9\pi} = \frac{1}{3\pi} \approx 0.1061 \text{ m/min}$$ **Final answer:** The water level in the cylindrical tank is rising at approximately $0.106$ meters per minute.